For the given reaction {\H{\C}_{|}_{CH_3}= {CHBr}{ ─────────\xrightarrow[(ii) \Red \hot \iron \tube \iron, 873 {K]{}}^{(i) {\NaNH}_{2}}(\A)_{\Major \product} [26 Feb 2021 Shift 1]

{CH}_{3} {CH}_{2} {CH}_{2} {NH}_{2}

{CH}_{|}_{CH_2}= {CH}- {NH}_{2}

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Haloalkanes and Haloarenes Part 1

Section: Chemistry

Question : 27 of 100

Marks: +1, -0

{\H{\C}↙{|}↙{CH_3}= {CHBr}{ ────────────────▶}↙{(ii) \Red \hot \iron \tube \iron, 873 {K}}↖{(i) {\NaNH}_{2}}(\A)↙{\Major \product}

[26 Feb 2021 Shift 1]

${CH}_{3} {CH}_{2} {CH}_{2} {NH}_{2}$
${CH}↙{|}↙{CH_2}= {CH}- {NH}_{2}$

Validate

Solution:

When 1-bromo-prop-1-ene reacts with

{NaNH}_{2}

(a strong base),

β

-elimination or dehydrobromination,

(- {HBr})

takes place to give propyne in next step. When propyne is passed through red hot iron tube, cyclic trimerisation (aromatisation also) occurs to produce mesitylene

(A)

as major product.

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