Question

How do you explain the following ?
(a) Presence of an aldehyde group in glucose
(b) Presence of a primary alcoholic group in glucose

Hint:

Aldehyde → oxime/cyanohydrin/Tollens'; primary −OH → HNO₃ gives saccharic acid.

Answer 1

(a) answer: First let us see what isomers $\mathrm{C_4H_9Br}$ can have. The carbon chain can be arranged in a few ways, and one of them is the branched one where bromine sits on the central carbon that is joined to three methyl groups. That is 2-bromo-2-methylpropane, $\mathrm{(CH_3)_3C\text{-}Br}$, also called tert-butyl bromide.

An $\mathrm{S_N1}$ reaction goes in two steps. In the first step the C to Br bond breaks on its own and a carbocation (a carbon with a positive charge) is formed. The faster and easier this carbocation forms, the faster the whole $\mathrm{S_N1}$ reaction. So we want the isomer that gives the most stable carbocation. When tert-butyl bromide loses its bromide, it forms $\mathrm{(CH_3)_3C^+}$, a tertiary carbocation. The three methyl groups around the positive carbon push electron density toward it and spread out the charge, which makes this carbocation very stable. Because its carbocation is the most stable of all the isomers, 2-bromo-2-methylpropane is the most reactive towards $\mathrm{S_N1}$.

(b) answer: This part is about the Wurtz reaction. When an alkyl halide is treated with sodium metal in dry ether, two alkyl groups join end to end and the two bromine atoms leave as $\mathrm{NaBr}$. So the product has double the carbon skeleton of the starting halide. We are told the product is 2,5-dimethylhexane. If we mentally cut this product in the middle, each half is a four-carbon group, $\mathrm{(CH_3)_2CHCH_2\text{-}}$, which is the isobutyl group. So the starting halide must be 1-bromo-2-methylpropane, $\mathrm{(CH_3)_2CHCH_2Br}$ (isobutyl bromide). Two of these couple together: \[ 2(CH_3)_2CHCH_2Br + 2Na \rightarrow (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaBr \]