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Chapter 22: Problem 517

Name the following structures: (a) \(\mathrm{CH}_{3}-\mathrm{NH}-\mathrm{C}_{2} \mathrm{H}_{5} ;\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{NH}-\mathrm{C}_{2} \mathrm{H}_{5}\) (c) \(\mathrm{NH}_{2}-\left(\mathrm{CH}_{2}\right)_{3}-\mathrm{NH}_{2}\) (d) \(\left(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\right)_{3} \mathrm{~N}\) (e) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{NH}_{2} ;\) (f) \(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}\) (g) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{NCH}_{3}\); (h) \(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{NH}-\mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5}\)

Short Answer

Expert verified
The short names for the given molecular structures are: (a) N-methylethanamine (b) N-phenylethanamine (c) 1,3-propanediamine (d) N,N,N-tripropylamine (e) 2-methylpropan-1-amine (f) 2-aminoethanoic acid (glycine) (g) N,N-diphenylmethanamine (h) N-phenylethanamine

Step by step solution

01

(a) CH3-NH-C2H5

This structure represents N-methylethanamine. The base structure is ethanamine (C2H5NH2), where a hydrogen atom of the amine group (-NH2) is replaced by a methyl group (CH3). Therefore, we write it as N-methylethanamine.
02

(b) C6H5-NH-C2H5

This structure represents N-phenylethanamine. The base structure is ethanamine (C2H5NH2) with one hydrogen (attached to Nitrogen) replaced by a phenyl group (C6H5). We follow a similar naming pattern as before - N-phenylethanamine.
03

(c) NH2-(CH2)3-NH2

This molecule is a diamine, as it has two amine groups (-NH2) in its structure. The chain contains 3 methylene groups (CH2), so the base structure is propane (C3H8). Because two hydrogens are replaced by amine groups, the name is following IUPAC nomenclature, 1,3-propanediamine.
04

(d) (CH3-CH2-CH2)3N

This molecule is a tertiary amine with a nitrogen atom bonded to three propyl groups (CH3CH2CH2). The name is following IUPAC nomenclature, N,N,N-tripropylamine.
05

(e) (CH3)2CH-NH2

This structure represents 2-methylpropan-1-amine. The base structure is propane (C3H8), where one hydrogen is replaced with an amine group, creating the base propylamine (C3H7NH2). The second carbon from the amine functional group also has a methyl group on it, so we name the compound: 2-methylpropan-1-amine.
06

(f) NH2-CH2-COOH

This molecule has an amine group (-NH2) and a carboxylic acid group (-COOH) connected by a methylene group (CH2). It is an α-amino acid, and in IUPAC nomenclature, it is referred to as 2-aminoethanoic acid. However, common nomenclature calls this molecule glycine.
07

(g) (C6H5)2NCH3

This structure represents N, N-diphenylmethanamine. The base structure is methanamine (CH3NH2), where both hydrogens attached to the nitrogen atom are replaced by two phenyl groups (C6H5). Using IUPAC nomenclature, we name the molecule N, N-diphenylmethanamine.
08

(h) C2H5-NH-CH2C6H5

This structure represents N-phenylethanamine in reverse form. The base compound is ethanamine (C2H5NH2) where a hydrogen atom of the amine group (-NH2) is replaced by a phenyl group (C6H5). Therefore, we write it as N-phenylethanamine.

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Most popular questions from this chapter

Compound A, C \(_{12} \mathrm{H}_{28} \mathrm{~N}_{2}\), was soluble in dilute acid, but did not react with benzenesulfony1 chloride. Upon addition of excess methyl iodide, Compound \(\mathrm{A}\) reacted. It was then treated with silver oxide, and the resulting product heated. Thus \(\mathrm{B}, \mathrm{C}_{8} \mathrm{H}_{14}\), was formed. B was treated with ozone and the ozonide decomposed with dilute acid in the presence of zinc dust. One of the two products isolates was

Show how the following transformations might be achieved with the aid of azide derivatives. (a) \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{OH} \rightarrow \mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\)

Explain why in the Ritter reaction with hydrogen cyanide and t-butyl alcohol, the product is t-butylamine and not t-buty1 cyanide (or trimethylacetic acid).

Suggest a means of demonstrating that benzylmethylphenyl-amine is not configurationally stable at room temperature.

Write balanced equations, naming all organic products, for the following reactions: (a) n-butyryl chloride + methylamine (b) acetic anhydride \(+\mathrm{N}\) -methylaniline (c) tetra-n-propylammonium hydroxide + heat (d) tetramethylammonium hydroxide + heat (e) N,N-dimethylacetamide + boiling dilute \(\mathrm{HCl}\) (f) benzanilide + boiling aqueous \(\mathrm{NaOH}\) (g) \(\mathrm{m}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{NHCH}_{3}+\mathrm{NaNO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (h) \(\mathrm{m}\) -toluidine \(+\mathrm{Br}_{2}(\mathrm{aq})\) in excess (i) \(\mathrm{p}\) -toluidine \(+\mathrm{NaNO}_{2}+\mathrm{HCl}\) (1) \(\mathrm{p}-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NHCOCH}_{3}+\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (k) benzanilide \(+\mathrm{Br}_{2}+\mathrm{Fe}\)
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