Match the infrared spectrum characteristics of 1 through 5 below with a compound

Question

Match the infrared spectrum characteristics of 1 through 5 below with a compound from the group which follows: Strong absorption band at 1715 cm^-1; no absorption band >3000 cm^-1 Sharp absorption bands at 3050 cm^-1 and 2120 cm^-1. Absorption bands at 3050 cm^-1, 1705 cm^-1, and 1630 cm^-1. A broad, strong absorption at 3350 cm^-1; no absorption from 1650 cm^-1 to 1780 cm^-1. A broad strong absorption from 2600 to 3400 cm^-1 and a strong band at 1710 cm^-1. A moderate, narrow band at 3310 cm^-1 and weak sharp band at 2030 cm^-1. A strong, somewhat broad band at 1090 cm^-1. Weak bands at 2710 and 2800 cm^-1 and a strong absorption at 1725 cm^-1. CH_3CH_2CCH=CH_2 CH_3CH_2CH_2C=CH -CHO CH_3CH_2CH_2CH(C_2H_5)CH_2CH_2CH_3 CH_3CH_2CH_2COOH (CH_3)_3CCH_2OH CH_3-O-CH_2CH_2CH_3 CH_2 = CHCH_2C = N

Explanation / Answer

This IR-spectroscopy..

1. Strong at 1715; meaning it has C=O bond .It is also common for a cyclic 6-carbon member ring...

If no previous data for 3000 HZ, then we are sure this is a KETONE and not an aldehyde.

This must be "E" since it has the 6 carbon rings, and the ketone group

2. 3050 --> typical for an alkene, that is C=C bond

2120 --> pretty near to a CN bond (cyanide)

therefore, th eony substance with nitrogen presence is "I"

3.

3050 --> typical for an alkene, that is C=C bond

1705 ---> C=O carbonyl group presence, either aldehyde or ketone

1630 --> double bond presence

Most likely "A" since it has a carbnyl group, and double bonds

4.

If

3050 --> presence of -OH group

if no sign in 1650 and 1780, no double bonds...

this must be an alcohol

Choose "G"

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