Q3. By how many degrees is the C–O–C angle of methoxymethane wider than the C–O–H angle of methanol (NCERT Fig. 7.1)?
111.7° − 108.9° = 2.8° (NCERT page 198).
Q4. Write methanol’s C–O–H angle, 108.9°, in degrees and minutes, and compare it with 109° 28′.
0.9° × 60 = 54′, so 108.9° = 108° 54′, which is 34′ less than 109° 28′ (reasoned from NCERT pages 198–199).
Q5. Would you expect the C–O–C angle of an ether with two tert-butyl groups to be larger or smaller than that of methoxymethane? Why?
Larger: NCERT puts the wide ether angle down to repulsion between the bulky alkyl groups (page 199), and tert-butyl groups are far bulkier than methyl groups (reasoned).