Messages posted by ashish.alias.ashu@gmail.com

I think the answer is (C). If it is correct, then I'll give you the solution (if required).

With due respect, sir I did not have the intention to hurt you or something like that.

This question was asked in one of the online test series.

I think that the options given to the above question are incorrect. Obvoiusly, since it is a single choice correct question, so three of them have to be incorrect. But what I think - One of the options should have been the one that the particle is moving towards y=2 (not along y=2 which is currently said by option (d) ). Sir if we can change the last option to that, then will the solution be fine???

OR

Is the given question correct, i.e., we don't have to change anything???

If the question is correct, then is the answer (d) or slightly modified one which should say that the particle moves towards y=2.

Really I thank you for your efforts in helping us out. Asking other experts from my side did not at all mean that I disagree with the solution provided by you. Instead I was just asking someone who may have different ideas or solution to share them.

There was this communication gap between us.

Thanks a lot Sir for your kind and helping nature, and continuing to post replies even when this thread continued for so long.

I have understood your answer.

But Sir, clearly mention whether the answer is (d) OR the answer should be that the particle will move towards y=2 (or z=2), not along y=2.

As fas as I have understood from your reply, the particle will move towards y=2 (or we can say that it'll move along the line z=2, which is the tangent at that point).

Looking forward to your reply!!!!!

I'm not saying that the question is incorrect. But from your reply and my understanding of the problem, I doubt the correctness of the option (d).

Other experts, please help us find the correct answer.

Someone please reply!!!

Sir, earlier we were not confused. The confusion started as this thread continued.

Sir, the charge will move towards the line y=2, or one can say that it'll move along the line z=2.

From the very begining, you are saying that option (d) is correct. How can that be so?

Option (d) says that the charge will move along the line y=2. It does not say that the charge will move towards the y=2. It is clearly written it will move ALONG not TOWARDS y=2.

Hence option (d) cannot be correct.

I think there is an error in the options given in the question.

We had already arrived at the conclusion that we have drawn now, i.e. it will move along z=2, or towards y=2.

As you said in your last reply, "both are same thing". But unfortunately there lies a huge difference between the two statements.

Guys, do it now. The process is very fast now. It will take hardly a minute for the whole process.

By the way, I had reserved the date yesterday only, at around 7.30 pm.

Thanks a lot guys!!!

Sir, see the figure that I have drawn:-

Even if we consider this in 3 dimensions, then how is the answer y=2 ?????

Explain with the help of this figure.

Aakash sir, please explain your answer in a liitle more detail. Since you have agreed that tangent at that point is z=2, then how is the answer y=2?

If you can explain it with a diagram, it'll be great help.

This thread has coninued for so long. I am really sorry for that.

Sir, please help.

Explain the answer in as much simple language and vast detail as possible, so as not to have the doubt again.

Really looking forward to your answer.

Someone please reply!!!

Assertion: $log\;2\;=\;1/(1.2)\;+\;1/(3.4)\;+\;1/(5.6)\;+\;............\;to\;infinity$

Reason: $\sum_{n=0}^{infinity}\;{n}^{2}/n!\;\;=\;2e$

(A) If both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) If both Assertion and Reason are true and but Reason is not the correct explanation for Assertion.
(C) If Assertion is true but the Reason is false.
(D) If Assertion is false but the Reason is true.

The answer given is (B).

Please provide the explanation for the Reason part.

Don't just quote the answer, give the explanation also.

Around 270-280 marks will take you in the first 1000.

This is because there is no element of symmetry in the compound formed. And both the mirror images are formed in equal amounts, so it gives a racemic mixture.

Yes, the answer to this question is 3.) (+ -) TRANS 1,2 DIBROMOCYCLOHEXANE

JPS, since we are carrying out the reaction with B2D6 , so don't you think CH3-CD=CHD would be formed, instead of the compound you have given. This is beacuse there is no source of hydrogen atoms. We are given with B2D6, and not B2H6.

If there are any errors in my answer, then please correct it.

Swaprules, the answer that you have given would be correct if instead of Acetic acid, we use . If that would have been the case, then also CH3-CD2-CHO would have formed.

Yes the answer to your previous question is (CH3)2-CH=CH2.

The reason for this answer is:- Since the halide given is tertiary, so instead of substitution, elimination reaction by the base takes place.

The answer to your other question is:- CH3-CD=CHD

The following figure is in Y-Z plane.

The particle is at y=0,z=2.

Since the point lies on the circle, the tangent at that point will be along y-axis (as shown by the dark black line)- Agreed!!!

So then the equation of the tangent should be z=2, and hence the particle should move along the line z=2. But unfortunately there is no such option given.

Also Aakash sir, you said that the particle will move along the line y=2, but sir the equation of the tangent at the given point is z=2. Please explain the dilemma.

Please reply.

Someone please reply!!!

I do not know in what context you are talking about the +I effect. If you are talking about the carbon atom of the -COOH group, it has almost zero +I effect in this case because that carbo atom is attached with two oxygen atoms (which are strongly electronegative). Hence they decrease the electron density of the carbon atom to such an extent that its +I effect has no meaning.

Benzoic acid is acidic due to the dominance of the resonance stabilization through the carboxylate anion formed after the removal of H+ ion. Both the canonical forms in this case are equally stable, leading to more resonance energy. Further, since rotation along the single bond is free, as soon as the caboxylate pi bond comes in the plane of the benzene ring, it is also stabilized by the -R effect of the benzene ring.

Thus there are two stability factors. +I effect has to take the back seat here.

Also, +I effect destabilizes the anion formed thereby decreasing the acidic character. It is the -I effect that increases the acidity, not the +I effect. -I effect decreases the negative charge density on the anion formed after the removal of the H+ ion, thereby stabilizing the anion formed.

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