Our teacher provides us extra questions for pratice, so one of the question was :->
'Find number of products if following are reacted with $NH_2OH$(hydroxyl amine) Cyclohex - (i) 1,4-dione (ii)1,3-dione (iii)1,2-dione (iv)1,2,3 trione (v)1,2,4-trione (vi)1,3,5-trione and (vii) cyclobut-1,2,3,4-tetraone.'
To solving this our teacher told us replace each =O group by 1$-CH_3$ and 1-H , then find the number of stereoisomer (including enantiomers) of the resultant compound which is the answer.(if you actually count it's tough to visualize the lone pair and -OH bond in the oximes group produced and gets long).
I have crosschecked , the answers pretty much matches.So my doubt is , is the short trick always right and if yes then please prove it, I would be happy if you can provide any links if the question already been answered.
I had another doubt, about another short trick my teacher told us about. Lets say there's a compound with 'n' carbons forming a closed cylic chain of 'n'carbon.At each carbon out of 2 -H , 1 -H is replaced by -Cl(chloride). The trick if n is odd ,the number of geometrical isomer of the compound is $2^{(\frac{n-1}{2})}$ and if n is odd then the number of geometrical isomers is $2^{(\frac{n}{2})}$.Pease kindly prove this also .
I know it should be asked in chemistry stack excahnge, but these tricks can't be proven without involving maths, Thanks in advance.
This is a partial answer only.
I think the first point Iabout replacing =O by -H and -CH3 provided at least one of the other bonds is -H} is more about chemistry than math.
Your second point about rings of C atoms is a math question! You have a ring of $n$ C atoms, each with a single bond to its two neighbours and -H and -Cl for the other two bonds. How many distinct structures can be formed?
If we orient the ring so that it is horizontal, each Cl can be above or below the ring. That is $2^n$ possibilities. But we are free to rotate the ring and also to flip it about a diameter, so many of the possibilities turn out to be physically indistinguishable.
Denote a Cl above the ring by U and below by D. It is not hard to see that for $n=4$ there are just four possibilities: UUUU, UUUD, UUDD and UDUD. For speed of typing I am writing each arrangement as a line rather than a ring. Rotating the first makes no difference, but flipping it gives DDDD (so 2 possibilities in all). Similarly, the second gives UUDU, UDUU, DUUU, DDDU, DDUD, DUDD, UDDD (so 8). The third gives UDDU, DDUU, DUUD (so 4) and the fourth gives DUDU (so 2). A total of 16. So $$4=2^{\lfloor\frac{n}{2}\rfloor}$$ as you suggested.
But $n=6$ does not support that formula. With some difficulty (perhaps) you can find there are 9 possibilities, not 8 as you might expect from $2^{6/2}$:
$UUUUUU;\ UUUUUD;\ UUUUDD$
$UUUDUD;\ UUDUUD;\ UUUDDD$
$UUDUDD;\ UUDDUD;\ UDUDUD$
The slightly tricky one is $UUDUDD$. If you flip it, you get the same arrangement rotated, rather than the mirror image. So $UUDUDD$ and $UUDDUD$ are not the same.
Counting these kind of things is notoriously error prone. The usual approach is known as Burnside's Lemma or Polya's Enumeration Theorem.
This example is closely related to the necklace problem or, more accurately, the bracelet problem. But I am not sure it is quite the same because of the way U changes to D when the ring is flipped. But if you are interested in pursuing it further, maybe I or someone on MSE can help you.