Question

Organic chemistry 2 question
I need help with problems #1 and #2, please showed works and neat writing, thank you

Answer 1

As per rules, only the first four will be answered.

1.

1. CH3-C-CH2-CH2-CH2-CH2-CH3 1 2 3 4 5 6 7 Carbon chain is of 7 carbons hence heptane. Ketone is at C-2. For ketone we use suffix 'one'. The 'e' of heptane is changed with 'one' and the ketone position is written at the beginning. So IUPAC name is 2-heptanone. 2. PhC-Ph Two phenyl groups attached to methyl ketone. Hence IUPAC name is diphenylmethanone. 3. Ph-CH=CH-CHO 3 2 1 Carbon chain consists of 3 carbons hence propane. Phenyl ring attached to C-3 and double bond at C-2. For double bond we use 'ene'. For aldehydes the suffix is 'al'. So IUPAC name 3-phenyl-prop2-enal. 4. CH3CH2CCH2CHO 5 4 3 2 1 Carbon chain consists of 5 carbons hence pentane. Here both ketone and aldehyde groups are present. But aldehydes get a higher priority over ketones. Hence aldehydes are written as suffix (al) and ketones as prefix (oxo) with the carbon positions attached. Hence the IUPAC name is 3-oxo-pentanal.