--%>

haloalkanes

define primary secondary and tertiary alkyl halides with examples

   Related Questions in Chemistry

  • Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Molecular basis of third law. The

    The molecular, or statistical, basis of the third law can be seen by investigating S = k in W.The molecular deductions of the preceding sections have led to the same conclusions as that stated in the third law of thermodynamics, namely, that a value can be

  • Q : Molarity what is the molarity of the

    what is the molarity of the solution prepared by dissolving 75.5 g of pure KOH in 540 ml of solution

  • Q : Why aryl halides are less reactive?

    Aryl halides are much less reactive towards nucleophilic substitution reactions than haloalkanes. The less reactivity of aryl halides can be described

  • Q : Formula of diesel Write a short note on

    Write a short note on the formula of diesel, petrol and also CNG?

  • Q : Strength of dilute acid of Sulfuric acid

    Select the right answer of the question.10ml of conc.H2SO4 (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be: (a)0.18 N (b)0.09 N (c) 0.36 N (d)1800 N

  • Q : Chemistry brief discription of relative

    brief discription of relative lowering of vapour pressure

  • Q : Volume hydrogen peroxide Choose the

    Choose the right answer from following. The normality of 10 lit. volume hydrogen peroxide is: (a) 0.176 (b) 3.52 (c) 1.78 (d) 0.88 (e)17.8

  • Q : Question based on vapour pressure While

    While a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in: (a) An increase in the boiling point of the solution (b) A decrease in the boiling point of solvent (c) The solution having a higher freezing point than

  • Q : Explain Second Order Rate Equations.

    Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o