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Question 60 of 69

Q.Identify A and B in the following sequence of reactions. CH3−Br→NaN3A→LiAlH4B+N2CH_3-Br \xrightarrow{NaN_3} A \xrightarrow{LiAlH_4} B + N_2

Puducherry TnboardTamil Nadu HSC (DGE) Board 2022Subjective· 2mImportance★★★★★
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AA is methyl azide (CH3N3CH_3N_3), formed by SN2S_N2 substitution of bromide by azide ion; BB is methylamine (CH3NH2CH_3NH_2), formed by LiAlH4LiAlH_4 reduction of the azide with loss of N2N_2.

Step 1: CH3−Br→NaN3ACH_3-Br \xrightarrow{NaN_3} A

Sodium azide (NaN3NaN_3) provides the azide ion (N3−N_3^-), a good nucleophile, which displaces the bromide leaving group from methyl bromide in an SN2S_N2 nucleophilic substitution:

CH3−Br+N3−→CH3−N3+Br−CH_3-Br + N_3^- \rightarrow CH_3-N_3 + Br^-

So AA is methyl azide, CH3−N3CH_3-N_3 (also written CH3N3CH_3N_3).

Step 2: A→LiAlH4B+N2A \xrightarrow{LiAlH_4} B + N_2

Lithium aluminium hydride is a strong reducing agent that reduces alkyl azides to primary amines. The azide group (−N3-N_3, containing 3 nitrogen atoms) is reduced so that only one nitrogen is retained (as the −NH2-NH_2 group of the amine), while the other two nitrogen atoms are extruded as nitrogen gas:

CH3−N3→LiAlH4CH3−NH2+N2↑CH_3-N_3 \xrightarrow{LiAlH_4} CH_3-NH_2 + N_2\uparrow

So BB is methylamine, CH3−NH2CH_3-NH_2.

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