Answer Q.No.40:- In IR spectrum the unknown molecule having molecular formula C

Question

Answer Q.No.40:-

In IR spectrum the unknown molecule having molecular formula C4H8O shows no absorption at 1630-1780 cm-1 or at 3200-3500 cm-1.

So the molecule will be ether group and gives IR absorption peak at 1300-1000 cm-1 (for C-O)

So the probable structure of the compound will be CH3-CH2-CO-CH3

Reason: for ketone the Carbonyl group (C=O) gives IR absorption peak at 1715-1710 cm-1

   for aldehyde the carbonyl group (C=O) gives IR absorption peak at 1725-1720 cm-1

for Alcohol the hydroxyl group (OH) gives IR absorption peak at 3500-3200 cm-1

for Carboxylic acid the OH group gives IR absorption peak at 3400-3200 cm-1

& for C=O gives IR absorption peak at 1710 cm-1.

For the above reason it can be seen that the unknown compound is belong to ether group.

Answer to question no.41

The compound I gives two peaks in 1HNMR, one peak for two methyl group and another peak for four rinh proton and the ratio is 6:4 because two equivalent methyl group has 6 equivalent proton and other 4 equivalent benzene proton has 4 equilvalent proton.

so answer will be I

Explanation / Answer

40. An formula of CHI0 a orgasic molecule having the mclecular mio snows no absorptions at 1530-1780rlorat 3200-3500 crrTo what class of compounds does the unikcnown beiong? A) alcohol B) carboxylic acid C) ketone D) ether E) aldehyde 41. Which of the following molecules will show two peaks with an ares ratio of 6:4 in 1 NMR spectrum? OCH CH3 CH3 CH Tv CH3 A) I B) II c) I

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