Answer to Question #71270 in Organic Chemistry for umesh rai




 WHY DOES ONLY HBr give anti anrkownikovs product in presence of per oxide among all the halo acid
Because only HBr produces a Br-radical (in presence of peroxide), which then reacts with the double bond. It is considered that HF and HCl are too stable to produce free radicals. HI produces I-radicals, but they have not enough activity for starting of radical chain reaction.  
This is also known as “the peroxide effect” or “the Kharasch effect”.
a) he helium (he) is an element by itself. he atom can only exhibit dispersion forces.b) hclh          1a                   1 × 1 e– = 1 e–cl         7a                   1 × 7 e– = 7 e–   total: 8 valence e–[readmore]cl is more electronegative than h, so hcl is a polar covalent compound. hcl exhibits dipole-dipole interaction.c) h2oh          1a                   1 × 2 e– = 2 e–o          6a                   1 × 6 e– = 6 e–   total: 8 valence e– since hydrogen is bonded directly to oxygen.h2o exhibits hydrogen bonding.d) cl2cl         7a                   1 × 7 e– = 14 valence e–composed of the same atom. there's no electronegativity difference.cl2 can only exhibit dispersion forces. ch3nh2 will appear as a stronger base since n for (ch3)2nh has bulkier substituents that will block n from accepting extra h (bl base = proton acceptor)[readmore]since ch3nh2  is the stronger base than (ch3)2nh, its c.acid will be weaker compared to (ch3)2nhstronger base → ch3nh2 stronger c.acid → (ch3)2nh3+ reno, please don't answer organic chemistry questions if you don't know the subject matter.
 
mixing acetic acid with sodium bicarbonate doesn't yield "gaseous water." also, you didn't actually answer the question and decided instead to list the first thing you found in wikipedia about a dangerous reaction that has nothing to do with the fact that acetic acid and hcl react the same way because they are both stronger acids than the conjugate acid of bicarbonate. use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4 because of the small size of boron atom and presence of only six electrons in its valence shell, b(oh)3 accepts a pair of electrons from oh- ion of h2o, releasing a proton.the reaction mechanism is: since the solution is a buffer, we can use the henderson-hasselbalch equation to calculate the ratio of [so32-]/[hso3-] to have a ph of 7.39:ph=pka+logconjugate baseweak acidin the solution, identify the weak acid and the conjugate base. recall from the bronsted-lowry definition that an acid is a proton (h+) donor and the base is a proton (h+) acceptor.• nahso3 is a salt that will give hso3- when it dissociatesnahso3 → na+ + hso3-• na2so3 is a salt that will give so32- when it dissociatesna2so3 → 2 na+ + so32-•hso3- is a weak acid.          - concentration of hso3- = concentration of nahso3• when hso3- loses h+ to give a conjugate base, so32- is formedhso32- - h+ → so32-- concentration of so32- = concentration of na2so3      hso3-    +   h2o     →        so32-        +       h3o+(weak acid)    (base)      (conjugate base)  (conjugate acid)we can rewrite the henderson-hasselbalch equation as:ph=pka+logso32-hso3-calculate [so32-]/[hso3-] ratio:• ph = 7.39• pka = 7.19ph=pka+logso32-hso3-7.39=7.19+logso32-hso3-logso32-hso3-=7.39-7.19logso32-hso3-log=0.2loglog has a base of 10 and when you divide a number by log you will take 10 to the power of the numberso32-hso3-=100.2[so32-]/[hso3-] = 1.58the ratio [so32-]/[hso3-] is 1.58 hclo3 > hclo2 > hclo > hbro we’re being asked to calculate the ph of 0.03m hc9h7o4.remember that weak acids partially dissociate in water and that acids donate h+ to the base (water in this case). the dissociation of hc9h7o4 is as follows:hc9h7o4 (aq) + h2o(l) ⇌ h3o+(aq) + c9h7o4– (aq)from this, we can construct an ice table. remember that liquids are ignored in the ice table and ka expression.the ka expression for hc9h7o4  is:ka=productsreactants=[h3o+][a-][hc9h7o4]note that each concentration is raised by the stoichiometric coefficient: [hc9h7o4], [h3o+] and [c9h7o4–] are raised to 1.plugin the value of ka and the terms from the ice chart: ka=[h3o+][a-][hc9h7o4]3.6 × 10-4=[x][x][0.03 -x]check if x is negligible in the denominator: if hc9h7o4ka>500 then x is negligible in the denominatorhc9h7o4ka= 0.03 3.6 ×10-4 = 83.3 < 500since the ratio is less than 500, we can’t remove (–x) from the equation. we first need to manipulate the equation to get a quadratic equation:3.6×10-4 = xx0.03-x0.03-x3.6×10-4   =x21.08 ×10-5-3.6×10-4 x = x2x2 +1.08 ×10-5-3.6×10-4 x =0we will then use the quadratic formula to solve for x. the quadratic formula is:x = -b±b2-4ac2afrom the equation, we have a = 1, b = 1.08×10–5, and c = –3.6 × 10–4. solving for x, the equilibrium concentration of [h3o+] and [c9h7o4–]:x = -b±b2-4ac2ax = -1.08×10-3±1.08×10-32-41-3.6 ×10-421x = -1.08×10-3±0.037962the calculation then splits into two ways:x = -1.08x10-3+0.037962x = 0.01844                          x = -1.08x10-3-0.037962x = -0.03904we take the positive value of x since the concentration of h3o+ and a– must increase. this means [h3o+] = [a–] = x = 0.01844 mnow calculate the ph using the formula below:ph = -log[h3o+]ph = -log[0.01844]ph = 1.73the ph of 0.030 m of aspirin is 1.73. a) van der waals (present, all molecule/atoms contain it)[readmore](b) dipole-dipole (present, c=o creates a partial charge on c and o)(c) hydrogen bonding (not present, h-bonds are only present in compounds with h attached on o, n or f)(d) ion-ion (not present, only appears on compound with full charges) the percent ionization of nh3 will increase.