17.C
30.A
will try rest tomorrow![1]
14 Answers
114)c
Boiling point of a compound decreases with increase in branching as the surface area decreases...So the I should have highest b.p
Its reverse for M.P..Melting point increases with branching.
22)b?
11for 14) But here why is KMno4 not oxidising the alkane to acid....is it due to the fact that it is 3 degree??
also for that 22....
CETANE NUMBER
1-methyl napthalene...cetane no..==0
octane number
heptane ...octane number ==0
Arihant Organic chem..pg 230
so shud'nt it be a??
137) Will go with A.
14) C (Can't oxidise to acid, methyl groups aren't gud leavers, so it can fight with only the Hydrogen... also, KMnO4 inserts Oxygens wherever it can find the place..!!)
30) B (Secondary "H" wud be replaced...making the carbon have 4 diff. groups- H, Cl, Me, Et)
1yup i also give it : 7 : A)
because in 2 and 3 the S.A of the molecule being decresed the Vander Waal's forces are reduced .....so the M.P and B.P will be less for 2 and 3.
so ans : A
1from which chapter are these questions taken ????
Sandipan Please inform me .
1hey in 14 wont the answer be B as the tert hydroxide group will be unstable and prefer elimination
13And who will be reponsible fr the Elimination....?
None, not KMnO4, not unless we introduce a dehydrating agent.
& moreso, can't/don't we get tert-Hydroxides as products in reactions...?
Its just the KmnO4 action we need to look upon, rather than foreseeing the future fr future reactions.
39Since when did KMnO4 start reacting with tertiary stuff anyway..
13Oh!....but it does here :D
Kyunki only a tiny tert-Hydrogen is at stake, not a difficult-to-chase-out functional group. :P
