Q3. Write the IUPAC names of (i) CH3COCH(CH3)2, (ii) CH2=CHCH2CHO, (iii) CH3COCH2CH2COCH3.
(i) 3-Methylbutan-2-one: number from the end nearer C=O. (ii) But-3-enal: the CHO carbon is C1, so the C=C starts at C3. (iii) Hexane-2,5-dione: both carbonyl groups are counted, at C2 and C5 (reasoned from NCERT page 229).
Q4. Name the ketone C6H5COCH(CH3)2 in both systems.
The acyl group (CH3)2CHCO– is isobutyryl, as in isobutyraldehyde (NCERT Table 8.1, page 230), so the common name is isobutyrophenone, built like propiophenone (NCERT page 229). IUPAC: a three-carbon chain with the C=O and the phenyl group on C1 and a methyl group on C2, prefixes in alphabetical order: 2-methyl-1-phenylpropan-1-one (reasoned from NCERT page 229).
Q5. Why do simple aldehydes and ketones generally have larger dipole moments than comparable ethers?
The C=O π electrons are easily pulled towards oxygen, as the dipolar resonance structure >C+–O− shows, so the bond is strongly polarised. NCERT says carbonyl compounds “have substantial dipole moments” and, with the missing “more” supplied (see the slip note), are more polar than ethers (NCERT page 231). The total molecular dipole also depends on substituents and geometry, so compare similar molecules (reasoned).
Q6. What is the shape round the carbonyl carbon, and what are its bond angles?
Trigonal planar, from its sp2 hybridisation: the carbon and the three atoms on it lie in one plane, with bond angles of about 120° (NCERT page 231).
Q7. The common name of (CH3)2C=CHCOCH3 is mesityl oxide. Explain its IUPAC name, 4-methylpent-3-en-2-one.
The chain has five carbons, so pent-; it is numbered from the end nearer C=O, which is C2 (-2-one); the C=C starts at C3 (-3-en-); and the methyl group is on C4 (NCERT Table 8.1, page 230; reasoned).