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Hello everyone! In this lecture, I am Going to provide solution to one of the questions of JEE advanced 2019. This question is from the chapter Aldehyde and ketones. It is based on reaction routes Which are very common type of questions asked in passage type. Through this question, we will also learn about the approach that we should develop over a period of time to get acquainted with advanced level questions. Here we are given a molecular formula. And our task is to figure out the various products produced in each step by following a particular sequence. In this question, we will be applying the reactions of Grignard reagent, Protonation of alcohol, electro meric effect, Anti Markonikov addition reactions, Addition of hydrogen hallide, Free radical substitution reactions. TIME STAMPS 00:00 QUESTION 01:00 APPROACH 03:42 DOUBLE BOND EQUIVALENT 06:42 POSSIBILITY 1 CYCLOHEXANONE 16:52 SOLUTION 18:52 DISCUSSION 27:12 THANK YOU Please go through the lecture and let me know if you have any questions or explanations that you may need. Thank you so much for watching. Please consider subscribing the channel to support my work.Also Share among your friends who might be benefitted from this lectures. You can contact me at: [email protected] #jee #neet #mains #organicchemistry #pyqs #solution #carbocations #ncert #class11 #class12 #cbse #solved