Stability of carbocations with -M groups in beta position First part
The original question was to arrange the following compounds in the order of their rate of dehydration with conc. $\ce{H2SO4}$:-

(A) $\ce{CH3-CH(OH)-CH2-CH=O}$
(B) $\ce{CH3-CH(OH)-CH2-C(=O)-CH3}$
(C) $\ce{CH3-CH(OH)-CH2-C#N}$
(D) $\ce{CH3-CH(OH)-CH2-NO2}$

In each of the cases, a carbocation will be formed (+ve charge on the $2$nd carbon via E1 mechanism). Since all the groups(not talking about $\ce{-OH}$, because that’s the leaving group) are electron-withdrawing, the weaker the electron-withdrawing group, the higher the rate.
So, the order according to me should be B &gt; A &gt; C &gt; D. But the answer given is the exact opposite.
&quot;Edit- My teacher who gave me this question now told me that my answer is correct when the reaction takes place in acidic medium and reverses in basic medium.&quot;
Second part
Here is a similar question,

Arrange the following compounds in the order of their rate of dehydration with conc. $\ce{H2SO4}$:-
(A) $\ce{CH3-CH2-CH(OH)-NO2}$
(B) $\ce{CH3-CH(NO2)-CH2-OH}$
(C) $\ce{NO2-CH2-CH2-CH2-OH}$

Here, the answer given is C &gt; B &gt; A which I agree with because the further the $\ce{-OH}$ group is from the nitro group, the more stable will be the carbocation. But I feel it contradicts the first question.
Is my answer for the first question correct? If not, what is the correct reasoning?
These would be the arguments I propose under the assumption that the RDS approximation is valid within the scope of the question:
The order (B &gt; A &gt; C &gt; D)  was said to be correct for the conc. $\ce{H2SO4}$/ $\Delta$ case since there would be no effect of $\ce{CH-}$ acidity since the protonation of the $\ce{-OH}$ group takes place and so the carbocation stability takes precedence meaning that the stronger the -I group, the slower the rate since the transition barrier increases.
However, for the scenario where only heat is used in dehydration (I assume basic conditions here), the reaction proceeds via E1cB  since the two main requirements for E1cB is that there is a relatively acidic hydrogen and a poor leaving group. Here, the RDS of the reaction would be the $\ce{C-H}$ bond cleavage. This would be stabilised by the presence of a strong -I group since a carbanion is being generated here which is stabilised by the delocalisation of the electrons via the inductive effect.
So, for such a scenario the rate order would be D &gt; C &gt; A &gt; B.
The answers depend upon the medium as you can see.
This entire argument seems hand wavy with the assumption taken but otherwise as Martin pointed out, this would not be trivial and would require quantum modelling:

Protonation and deprotonation reactions in a catalytic cycle are notoriously hard to calculate. The RDS approximation will break down.