Why does heating glucose in the excess of HI produce n-hexane?

Asked Jan 15 '19 at 06:12

Active Jan 15 '19 at 06:25

Viewed 46 times

Alcohols react with acid halides such as $\ce{HI}$ to create alkyl halides, as such:

$$\ce{C2H5OH + HX → C2H5X}$$

However, this doesn't seem to be the case with glucose, as the entire chain is reduced to an alkane instead. Is this due to the steric hindrance factor of the large chain or perhaps the aldehydic group attached to it?

An explanation would be appreciated.

edited Jan 15 '19 at 06:25

andselisk

37,604
14
131
217

asked Jan 15 '19 at 06:12

Confused

1

Iodine is somewhat special among halogens, in that it falls off easily. – Ivan Neretin Jan 15 '19 at 06:22
@IvanNeretin Even though Iodine is a great leaving group, can it be replaced with a hydrogen? – Confused Jan 15 '19 at 07:29
Leaving is irrelevant; falling off is another story. – Ivan Neretin Jan 15 '19 at 07:40

Why does heating glucose in the excess of HI produce n-hexane?

0 Answers0