Free Essay


In: Science

Submitted By joemo18
Words 533
Pages 3
Name: Josiah Morales

Title: Amines

Amines are organic compounds containing nitrogen containing the functional group R−NH2. Amines are capable of forming hydrogen bonds and small amines are quite soluble in water. Amines are one of the primary functional groups in amino acids, and are often found as part of compounds that are physiologically active or used in medications. In this experiment, the identity of an unknown amine or amine salt had to be determined using Solubility testing and the Hinsberg test.

Apparatus and Materials: See Laboratory Manual, pages 40-42(Gary Gray. Laboratory Procedures: Organic Chemistry 2. Chemistry 3404. Wayland Baptist University, Plainview, TX.)

Procedure: See Laboratory Manual, pages 40-42(Gary Gray. Laboratory Procedures: Organic Chemistry 2. Chemistry 3404. Wayland Baptist University, Plainview, TX.)


Boiling point of unknown – 181 ͦC

Refractive index of unknown- 1.586

Solubility Test: Compound | Solubility | Odor | Aniline | - | Odor remained | Aniline hydrochloride | + | Odor formed then disappeared | Aniline sulfate | - | no odor | 4-Bromoaniline | - | Odor formed then disappeared | Methylamine hydrochloride | + | Odor remained | Pyridine | + | Odor remained | Unknown #4 | - | Odor remained |

Hinsberg Test: Compound | Reaction | Precipitate | Dissolved in Water | Aniline | - | - | - | N-Methylaniline | + | + | - | Triethylamine | + | - | - | Unknown Amine | - | - | - |

Infrared Spectrum of Unknown:

Discussion and Conclusion:
To determine the unknown amine, physical tests were done followed by two confirmatory test which were solubility and hinsberg respectively. The melting point was 181 ͦC and the refractive index was 1.586 indicating the unknown amine was aniline.
Most amines are soluble in water usually and dissolve in aqueous acid due to their ability to hydrogen bond. As the carbon chains get longer, solubility decreases as the hydrocarbon chains have to force their way between water molecules, breaking hydrogen bonds between water molecules. In the experiment the unknown was not soluble in water.
Secondly the Hinsberg test can distinguish primary, secondary, and tertiary amines and is based on sulfonamide formation. In the Hinsberg test, an amine is reacted with benzene sulfonyl chloride. If a product forms, the amine is either a primary or secondary amine, because tertiary amines do not form stable sulfonamides. If the sulfonamide that forms dissolves in aqueous sodium hydroxide solution, it is a primary amine. If the sulfonamide is insoluble in aqueous sodium hydroxide, it is a secondary amine. The sulfonamide of a primary amine is soluble in an aqueous base because it still possesses an acidic hydrogen on the nitrogen, which can be lost to form a sodium salt. The unknown did not react indicating it was a tertiary amine.
These results do not correspond with the compound aniline therefore there is some error in the experiment.

Problems: 1. Would you expect the reaction product from benzenesulfonyl chloride and ammonia to be soluble or insoluble in a basic solution? Explain.

The benzoyl chloride will react with the ammonia via a nucleophillic substitution reaction. The strong electron withdrawing sulphonyl group in benzene sulphonamide makes the hydrogen atom linked to nitrogen to become strongly acidic making it is soluble in a basic solution like KOH or NAOH.

2. How would you prepare aniline from aniline hydrochloride?

I would add a strong base such as NaOH or KOH to the aniline hydrochloride to obtain aniline.

3. Predict the NMR spectrum of 2-propanamine.

Similar Documents

Free Essay


...EPOXY RESIN A Technical Report Presented to Prof. Jaime Dilidili Department of Management College of Economics, Management and Development Studies Cavite State University Indang, Cavite In partial fulfillment of the requirements in CENG 28 (Materials of Engineering) ERJHON C. DINGLASAN October 2015 ABSTRACT Knowledge of the materials has always been necessary for those who design. With today’s stringent demand on quality regarding manufacture, materials and execution, it is often assumed that purchasers, architects and contractors posses the necessary knowledge of that with which they work. This paper described one of the commonly used engineering materials nowadays. Epoxy resin which is routinely used as adhesives, coating encapsulates, casting materials, potting compounds and binders. INTRODUCTION Epoxy resins were first commercialized in 1946 and are widely used in industry as protective coatings and for structural applications, such as laminates and composites, tooling, molding, casting, bonding and adhesives, and others. The ability of the epoxy ring to react with a variety of substrates gives the epoxy resins versatility. Treatment with curing agents gives insoluble and intractable thermoset polymers. Some of the characteristics of epoxy resins are high chemical and corrosion resistance, good mechanical and thermal properties, outstanding adhesion to various substrates, low shrinkage upon cure, good electrical insulating properties,......

Words: 6511 - Pages: 27

Free Essay

Copper-Catalysed Selective Hydroamination Reactions of Alkynes

... Buchwald* The development of selective reactions that utilize easily available and abundant precursors for the efficient synthesis of amines is a long-standing goal of chemical research. Despite the centrality of amines in a number of important research areas, including medicinal chemistry, total synthesis and materials science, a general, selective and step-efficient synthesis of amines is still needed. Here, we describe a set of mild catalytic conditions utilizing a single copper-based catalyst that enables the direct preparation of three distinct and important amine classes (enamines, α-chiral branched alkylamines and linear alkylamines) from readily available alkyne starting materials with high levels of chemo-, regio- and stereoselectivity. This methodology was applied to the asymmetric synthesis of rivastigmine and the formal synthesis of several other pharmaceutical agents, including duloxetine, atomoxetine, fluoxetine and tolterodine. C omplex organic molecules play a crucial role in the study and treatment of disease. The extent to which they can be utilized in these endeavours depends on the efficient and selective chemical methods for their construction1. Amines are widely represented in biologically active natural products and medicines2 (a small selection of which are shown in Fig. 1a). Consequently, the selective assembly of amines from readily available precursors is a prominent objective in chemical research3. There are a number of powerful methods that address...

Words: 6397 - Pages: 26

Free Essay


...CHAPTER 15 ORGANIC COMPOUNDS AND THE ATOMIC PROPERTIES OF CARBON 15.15 a) Octane denotes an eight carbon alkane chain. A methyl group (–CH3) is located at the second and third carbon position from the left. CH3 CH3 CH CH CH2 CH2 CH2 CH2 CH3 CH3 b) Cyclohexane denotes a six-carbon ring containing only single bonds. Numbering of the carbons on the ring could start at any point, but typically, numbering starts at the top carbon atom of the ring for convenience. The ethyl group (–CH2CH3) is located at position 1 and the methyl group is located at position 3. CH2 CH3 1 6 2 3 5 CH3 4 c) The longest continuous chain contains seven carbon atoms, so the root name is “hept-.” The molecule contains only single bonds, so the suffix is “-ane.” Numbering the carbon chain from the left results in side groups (methyl groups) at positions 3 and 4. Numbering the carbon chain from the other will result in side groups at positions 4 and 5. Since the goal is to obtain the lowest numbering position for a side group, the correct name is 3,4– dimethylheptane. Note that the prefix “di-” is used to denote that two methyl side groups are present in this molecule. d) At first glance, this molecule looks like a 4 carbon ring, but the two –CH3 groups mean that they cannot be bonded to each other. Instead, this molecule is a 4–carbon chain, with two methyl groups (dimethyl) located at the position 2 carbon. The correct name is......

Words: 1379 - Pages: 6

Premium Essay

Organic Chemisty

...Oxidation 13. Ozonolysis cleavage 14. Cationic polymerization 15. Free-radical polymerization 16. Addition of halogens and water (Halohydrin formation) Alkynes Preparation: 1. Dehydrohalogenation of vicinal and germinal dihalides 2. Dehalogenation of vicinal tetrahaloalkanes 3. Substitution Reaction: 4. Hydrogenation 5. Hydrohalogenation 6. Halogenation 7. Hydration (keto-enol tautomerization) 8. Reaction of acidic terminal hydrogen (acid-base reaction) Alkyl Halides Preparation: 1. Addition of halogen halide to alkenes 2. Reaction of phosphorus and sulfur halides with alcohols Reaction: 3. Hydrolysis 4. Williamson ether synthesis 5. Nitrile formation 6. Amine formation 7. Alkene formation 8. Grignard formation Phenols Preparation: 1. Pyrolysis of sodium benzene sulfonate 2. Dow process 3. Air oxidation of cumene Reaction: 4....

Words: 623 - Pages: 3

Free Essay


...STABILITY | Stable under ordinary conditions. | GENERAL DESCRIPTION & APPLICATIONS | Amide is a group of organic chemicals with the general formula RCO-NH2 in which a carbon atom is attached to oxygen in solid bond and also attached to an hydroxyl group, where 'R' groups range from hydrogen to various linear and ring structures or a compound with a metal replacing hydrogen in ammonia such as sodium amide, NaNH2. Amides are divided into subclasses according to the number of substituents on nitrogen. The primary amide is formed from by replacement of the carboxylic hydroxyl group by the NH2, amino group. An example is acetamide (acetic acid + amide). Amide is obtained by reaction of an acid chloride, acid anhydride, or ester with an amine. Amides are named with adding '-ic acid' or '-oic acid' from the name of the parent carboxylic acid and replacing it with the suffix 'amide'. Amide can be formed from ammonia (NH3). The secondary and tertiary amides are the compounds which one or both hydrogens in primary amides are replaced by other groups. The names of secondary and tertiary...

Words: 3103 - Pages: 13

Premium Essay

Macromolecules and Their Parts

...There are four macromolecules that make up all living things. These macromolecules are proteins, carbohydrates, lipids, and nucleic acids. All of these macromolecules are formed from functional groups. The functional group hydroxyl is found in every macromolecule. All of these macromolecules except for lipids are hydrophilic. All of these macromolecules are broken down into monomers by hydrolysis reactions. Proteins have methyl groups, amino groups, and carboxyl groups. Proteins are polymers of amino acids. Proteins functions are determined by how they are folded. Proteins with a primary structure have no function. Proteins with a secondary structure are in the shape of an alpha helix but still have no function. Once a protein achieves tertiary structure that protein has a function. Quaternary structures in proteins are made of many motifs and have the most advanced functions of all the structures, many of them being enzymes due to their three dimensional structure. However, a protein can become unfolded, and once it does it cannot be reversed, this is called denaturation and makes the protein loose its functions. All proteins and polypeptides are links of amino acids held together by peptide bonds. Peptide bonds are formed between two molecules when the carboxyl group and the amino groups in two atoms react which causes dehydration synthesis. Carbohydrates have carbonyl groups, and are organic compounds every time due to the presence of a carbon atom and also have......

Words: 491 - Pages: 2

Free Essay

Table of Ir Absoprtion

...Stretch Alkynyl C-H Stretch Alkynyl C=C Stretch Aromatic C-H Stretch Aromatic C-H Bending Aromatic C=C Bending Alcohol/Phenol O-H Stretch Carboxylic Acid O-H Stretch Amine N-H Stretch Nitrile C=N Stretch Aldehyde C=O Stretch Ketone C=O Stretch Ester C=O Stretch Carboxylic Acid C=O Stretch Amide C=O Stretch Amide N-H Stretch Characteristic Absorption(s)(cm-1) Notes Alkane C-H bonds are fairly ubiquitous and 2950 - 2850 (m or s) therefore usually less useful in determining structure. 3100 - 3010 (m) 1680 - 1620 (v) Absorption peaks above 3000 cm-1 are frequently diagnostic of unsaturation ~3300 (s) 2260 - 2100 (v) ~3030 (v) 860 - 680 (s) 1700 - 1500 (m,m) 3550 - 3200 (broad, s) 3000 - 2500 (broad, v) 3500 - 3300 (m) See "Free vs. Hyrdogen-Bonded Hydroxyl Groups" in the Introduction to IR Spectra for more information Primary amines produce two N-H stretch absorptions, secondary amides only one, and tetriary none. 2260 - 2220 (m) 1740 - 1690 (s) 1750 - 1680 (s) 1750 - 1735 (s) 1780 - 1710 (s) 1690 - 1630 (s) The carbonyl stretching absorption is one of the strongest IR absorptions, and is very useful in structure determination as one can determine both the number of carbonyl groups (assuming peaks do not overlap) but also an estimation of which types. 3700 - 3500 (m) As with amines, an amide produces zero to two NH absorptions depending on its type....

Words: 257 - Pages: 2

Free Essay

Characteristic Ir Frequencies

...|Characteristic IR Absorption Frequencies of Organic Functional Groups | | |Type of Vibration |Characteristic Absorptions |Intensity | |Functional Group | |(cm-1) | | |Alcohol | | |O-H |(stretch, H-bonded) |3200-3600 |strong, broad | |O-H |(stretch, free) |3500-3700 |strong, sharp | |C-O |(stretch) |1050-1150 |strong | |Alkane | | |C-H |stretch |2850-3000 |strong | |-C-H |bending |1350-1480 |variable | |Alkene | | |=C-H |stretch...

Words: 446 - Pages: 2

Free Essay

Nylon 6,6 water, HCl, or once in awhile NaCl. The simplest polycondensation for making nylons is the reaction of a diamine and a diacid: O O ║ ║ H2N-CH2-CH2-CH2-CH2-CH2-CH2-NH2 + HOC-CH2-CH2-CH2-CH2-COH ↔ This reaction might not normally go to high conversions: H O O │ ║ ║ ↔ H2N-CH2-CH2-CH2-CH2-CH2-CH2-N-C-CH2-CH2-CH2-CH2-COH + H20 Removing water (usually by carrying out the reaction under vacuum and/or at elevated temperature so the water evaporates), makes the reaction go to high conversion thanks to LeChatlier's Principle: H HO O H H O O H H O O │ │║ ║ │ │ ║ ║ │ │ ║ ║ -N-(CH2)6-N-C-(CH2)4-C-N-(CH2)6-N-C-(CH2)4-C-N-(CH2)6-N-C-(CH2)4-C- + H20 Because two different chemical were uses, one that has two amine groups and another that has two acid groups, the polymer can be represented by two different segments: -A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B-A-B- Making nylon is even easier if you use a...

Words: 1018 - Pages: 5

Free Essay

Dye Absorbtion of Fabrics

...Conversation opened. 1 read message. Skip to content Using Gmail with screen readers Move to Inbox More 1,859 of many (no subject) Inbox x a vaziri AttachmentsApr 29 to me Attachments area Click here to Reply or Forward 6.85 GB (45%) of 15 GB used Manage Terms - Privacy Last account activity: 0 minutes ago Details Get a Gmail custom address (you@yourcompany) Recommended for business email users Page 1 of 1 The wool absorbed the dye the strongest followed by the silk. Cotton and nylon absorbed the dye slightly giving a yellow color. Polyester did not absorb the dye well at all. Because of wools structure, which contains amine and ketone groups, it is very polar. It is also an amino acid chain (poly peptide). The polarity of the wool enables the methyl red (which is also very polar because of 1 carboxylic acid group, tertiary nitrogen, and a double bond nitrogen bridge that connects the two aromatic rings of the methyl red molecule) to bond with the wool very well. The biggest factor in which the dye bonds to the wool is that hydrogen bonds can form between the OH on the dye and the Nitrogens and Oxygens within the amino acid structure. Silk absorbed very well also. This is because the nitrogens and carbonyls within the polypeptides in the amino acid sequence of the silk fabric. This allows for the Hydrogen bonding between the OH of the dye and the carbonyl and Nitrogens within the polypeptide......

Words: 520 - Pages: 3

Free Essay

Functional Characterization of Chitin and Chitosan

...Current Chemical Biology, 2009, 3, 203-230 203 Functional Characterization of Chitin and Chitosan Inmaculada Aranaz, Marian Mengíbar, Ruth Harris, Inés Paños, Beatriz Miralles, Niuris Acosta, Gemma Galed and Ángeles Heras* Department of Physical Chemistry II, Faculty of Pharmacy, Institute of Biofunctional Studies, Complutense University, Paseo Juan XXIII, nº 1. Madrid 28040, Spain Abstract: Chitin and its deacetylated derivative chitosan are natural polymers composed of randomly distributed -(1-4)linked D-glucosamine (deacetylated unit) and N-acetyl-D-glucosamine (acetylated unit). Chitin is insoluble in aqueous media while chitosan is soluble in acidic conditions due to the free protonable amino groups present in the D-glucosamine units. Due to their natural origin, both chitin and chitosan can not be defined as a unique chemical structure but as a family of polymers which present a high variability in their chemical and physical properties. This variability is related not only to the origin of the samples but also to their method of preparation. Chitin and chitosan are used in fields as different as food, biomedicine and agriculture, among others. The success of chitin and chitosan in each of these specific applications is directly related to deep research into their physicochemical properties. In recent years, several reviews covering different aspects of the applications of chitin and chitosan have been published. However, these reviews have not taken into......

Words: 10560 - Pages: 43

Premium Essay

11300a Bioinformatics Homework

...Name: CHEN LIN Student ID: 44141569-3 11300A Bioinformatics Homework 2 Nov. 18, 2014 1. Sequence Alignment Using Dynamic Programming (DP) There are two amino acid sequences, seq1: COELACANTH and seq2: PELICAN. Obtain the global alignment by using DP (the Needleman-Wunsch algorithm) . $+ 1 for letter that match ! Scoring scheme : #- 1 for mismatches !- 1 for gaps "  Answer: ————————————————————————— C O E L A C A N T H P E L I C A N ————————————————————————— -1 -1 1 1 -1 1 1 1 -1 -1 λ λ P E L I C A N Seq1 Seq2 C -1 -1 -2 -3 -4 -3 -4 -5 O P O -2 -2 -2 -3 -4 -4 -4 -5 E E E -3 -3 -1 -2 -3 -4 -5 -5 L L L -4 -4 -2 0 -1 -2 -3 -4 A I A -5 -5 -3 -1 -1 -2 -1 -2 C C C -6 -6 -4 -2 -2 0 -1 -2 A A A -7 -7 -5 -3 -3 -1 1 0 N -8 -8 -6 -4 -4 -2 0 2 N N T -9 -9 -7 -5 -5 -3 -1 1 T — H -10 -10 -8 -6 -6 -4 -2 0 H — 0 -1 -2 -3 -4 -5 -6 -7 C — Name: CHEN LIN Student ID: 44141569-3 Seq1 Seq2 C P O — E E L L A I C C A A N N T — H — 2. In this question you will use two different dot matrix analysis servers to analyze the sequence of the human low density lipoprotein receptor (NP_000518). You will run a dot matrix analysis of this protein against itself (which means you will need to enter its sequence in both boxes on the website). A. First use Dottup ( Set the word size to 2 (“word size” is basically the same as “window”). Using a word size of 2, the algorithm will scan a window of 2......

Words: 486 - Pages: 2

Free Essay

Stardust Spacecraft

...Stardust is the NASA spacecraft that took samples of the Wild 2 comet. This craft returned samples that showed sampled of glycine, an amino acid used to make proteins. The mission took place in January of 2004 and the data was sent back to Earth in January 2006. These results were the first to prove that there are life ingredients in space. This proof restarts the issue of whether our species was created by a meteor or comet that came to Earth and left these similar samples. It’s perfectly logical to think believe if this one comet had these particles that many more could too. That they could be remnants of other planets or celestial bodies even. Dr. Carl Pilcher, Director of the NASA Astrobiology Institute which co-funded the research said, “The discovery of glycine in a comet supports the idea that the fundamental building blocks of life are prevalent in space, and strengthens the argument that life in the universe may be common rather than rare.” This is simply a hint of possibilities for what could be out there. Through this two-year journey much was accomplished. These accomplishments give scientists and astronomers even more of a reason to look into space in search for many things from life to more life ingredients. If our science today can accomplish something like this then the limitations are endless for future scientists and what they can find. Professor Donald E. Brownlee said it best when he said, “The discovery of amino acids in the returned comet......

Words: 263 - Pages: 2

Free Essay


...atter2/7/2014 Chapter 4: Carbon & Molecular Diversity of Life Chapter 4 Reading O Pages 58-61 O ½ page 63 (4.3 only) Organic Molecules O Contain carbon. Carbon! Carbon! Carbon! Testosterone DNA Methane Figure 4.2 Miller Experiment (1953) “Atmosphere” CH4 Electrode What about CO2 O Not traditionally considered organic. O It is the source of carbon for all other 2. Hydrogen gas (H2), ammonia (NH3), & water vapor in “atmosphere”. Water vapor 3. “Lightning”. Condenser Cooled “rain” containing organic molecules organic molecules!! 1. Salt (sea) water heated. Cold water H2O “sea” Sample for chemical analysis 4. “rain” condenses as cooled. Molecules collect in flask. 1 2/7/2014 Results of Miller Experiment O Found amino acids, hydrocarbons and other Carbon O The backbone of all ORGANIC molecules O 4 valence electrons O Forms up to four covalent bonds O Single or double bonds biological molecules that spontaneously formed, abiotically! O Versatility in attachable functional groups Carbon Skeleton Variations Hydrocarbons O Fats O Fossil fuels O C-H bonds store Nucleus Fat droplets Carbon skeletons vary in length Double bond Carbon skeletons may have double bonds, which can vary in location ENERGY! Carbon skeletons may be unbranched or branched Carbon skeletons may be arranged in rings 10 m (a) Part of a human adipose cell (b) A fat molecule O Insoluble (C-H......

Words: 304 - Pages: 2

Premium Essay

11300a Bioinformatics

...11300A Bioinformatics Homework 2 Nov. 17, 2015 1. Sequence Alignment Using Dynamic Programming (DP) There are two amino acid sequences, seq1: COELACANTH and seq2: PELICAN. Obtain the global alignment by using DP (the Needleman-Wunsch algorithm) . [pic] |λ |c |o |e |l |a |c |a |n |t |h | |λ |0 |-1 |-2 |-3 |-4 |-5 |-6 |-7 |-8 |-9 |-10 | |p |-1 |-1 |-2 |-3 |-4 |-5 |-6 |-7 |-8 |-9 |-10 | |e |-2 |-2 |-2 |-1 |-2 |-3 |-4 |-5 |-6 |-7 |-8 | |l |-3 |-3 |-3 |-2 |0 |-1 |-2 |-3 |-4 |-5 |-6 | |i |-4 |-4 |-4 |-3 |-1 |-1 |-2 |-3 |-4 |-5 |-6 | |c |-5 |-5 |-5 |-4 |-2 |-2 |0 |-1 |-2 |-3 |-4 | |a |-6 |-6 |-6 |-5 |-3 |-1 |-1 |1 |0 |-1 |-2 | |n |-7 |-7 |-7 |-6 |-4 |-2 |-2 |0 |2 |1 |0 | | 2. In this question you will use two different dot matrix analysis servers to analyze the sequence of the human low density lipoprotein receptor (NP_000518). You will run a dot matrix analysis of this protein against itself (which means you will need to enter its sequence in both boxes on the website). A. First use Dottup ( Set the word size to 2 (“word size” is basically the same as “window”). Using a word size of 2, the algorithm will scan a window of 2 amino acids and put one dot in the matrix when the two sequences have identical amino acids. Dottup has no threshold, so it is simpler than Dotmatcher. a) Print (or copy and paste) the output from Dottup and turn it in. B. Now use......

Words: 367 - Pages: 2