Chemistry assignment?

1. An optically active compound A(C10H17Br) when treated with alcoholic solution of KOH yields 2 compounds B and C of molecular formula C10H16 of which B is optically active. Ozonolysis of A followed by treatment with Zn-H2O gives acetone as one of the products. Hydrogenation of either B or C yields 4-isopropyl-1-methyl cyclohexane. Identify A,B,C.

2. Coniferyl alcohol(X), not soluble in water or NaHCO3 has molecular formula C10H12O3. A solution of Br2 in CCl4 is decolourised by X forming C10H12O3 Br2(A). Upon reductive ozonolysis of X, 4-hydroxy-3-methoxy benzaldehyde and B(C2H2O4) are produced. X reacts with C6H5COClin presence of a base to form C (C24H20O5)This product rapidly decolourises aqueous solution of KMnO4 and is insoluble in NaOH. X reacts with cold HBr to form D(C10H11O2Br). X reacts with HI to produce E(C9H9O2I)and CH3I. In aqueousbase,CH3I and X form F(C11H14O3), which is not soluble in strong base but decolourises aqueous solutionof Br2. Deduce X.

Answer:
1) A = 1-bromo- 4-isopropylidene- 1-methylcyclohexane
B = 4-Isopropylidene- 1-methylcyclohexene
C = 4-isopropylidene- 1-methylenecyclohexane

2) 4- (3-hydroxy- 1-propenyl)- 2-methoxyphenol

see structures here:
http://img329.imageshack.us/img329/8619/...

for structures of #2 A-F, see here:
http://img507.imageshack.us/img507/4351/...
i love this game.
A posesses a ispopropylidene group ( the acetone reaction). it also has at least one assymetric carbon. the core is therefore a 1-methyl-4isopropylidene-cyclo... which in order to be able to yield two elimination isomers, need to be 2-Br-substituted therefore A is
2-Bromo-4-isopropylidene-1-met...
B will then have to be with a double bond in the ring, allowing still for the asymmetrical C atom , a.k.a.
3-Isopropylidene-6-methyl-cycl...
and C is the other elimination product, symmetrical,(no asym carbons)
4-Isopropylidene-1-methyl-cycl...

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