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Predict the products of the following reactions:

(a) ±ð³æ³¦±ð²õ²õ â¶Ä‰N±á3 â¶Ä‰+ â¶Ä‰Ph−CH2CH2CH2Br→

(b)1−bromopentane→(2)LiAlH4(3)H3O+(1)NaN3

(c)

(d) product â¶Ä‰from â¶Ä‰part â¶Ä‰c→heat

(e)

(f) product â¶Ä‰from â¶Ä‰part â¶Ä‰e→(3)heat(1)excessCH3I(2)Ag2O

(g)

(h)

(i)

(j) product â¶Ä‰from â¶Ä‰part â¶Ä‰i→(2)H3O+(1)LiAlH4

(k)

(l)

Short Answer

Expert verified

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

(l)

Step by step solution

01

Prediction of products

(a) (3-bromopropyl)benzene on treatment with excess ammonia gives the product as 3-phenylpropan-1-amine.

reaction a

(b) 1-bromopentane on treatment with sodium azide forms 1-azidopentane, which on further reduction with lithium aluminum hydride gives the final product as pentan-1-amine.

reaction b

(c) 1-methylpiperidine reacts with hydrogen peroxide to give the product as 1-methylpiperidine 1-oxide.

reaction c

(d) 1-methylpiperidine 1-oxide on heating gives the product as N-methyl-N-(pent-4-en-1-yl)hydroxylamine.

reaction d

(e) Octahydro-1H-quinolizine on treatment with excess methyl iodide gives the initial product as 5-methyldecahydroquinolizin-5-ium, which on further reaction with silver oxide followed by heating gives the final product as 2-(but-3-en-1-yl)-1-methylpiperidine.

reaction e

(f) 2-(but-3-en-1-yl)-1-methylpiperidine on treatment with excess methyl iodide gives the initial product as 2-(but-3-en-1-yl)-1,1-dimethylpiperidin-1-ium, which on further reaction with silver oxide followed by heating gives the final product as N,N-dimethylnona-1,8-dien-5-amine.


reaction f

(g) Piperidine on treatment with sodium nitrite in presence of hydrochloric acid gives the product as 1-nitrosopiperidine.

reaction g

(h) Nitrobenzene on treatment with zinc in presence of hydrochloric acid gives the product as aniline.

reaction h

(i) 2-cyclohexylacetyl chloride on treatment with methylamine in presence of pyridine gives the product as 2-cyclohexyl-N-methylacetamide.

reaction i

(j) 2-cyclohexyl-N-methylacetamide on reduction with lithium aluminum hydride gives the final product as 2-cyclohexyl-N-methylethanamine.

reaction j

(k) (E)-N-(heptan-3-ylidene)methanamine on reduction with lithium aluminum hydride gives the final product as N-methylheptan-3-amine.

reaction k

(l) 2-methyl-3-phenylpropanenitrile on reduction with lithium aluminum hydride gives the final product as 2-methyl-3-phenylpropan-1-amine.

reaction l

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